Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $a = \dfrac{p^2 + 2p - 24}{4p + 32} \div \dfrac{5p - 20}{p + 8} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{p^2 + 2p - 24}{4p + 32} \times \dfrac{p + 8}{5p - 20} $ First factor the quadratic. $a = \dfrac{(p - 4)(p + 6)}{4p + 32} \times \dfrac{p + 8}{5p - 20} $ Then factor out any other terms. $a = \dfrac{(p - 4)(p + 6)}{4(p + 8)} \times \dfrac{p + 8}{5(p - 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (p - 4)(p + 6) \times (p + 8) } { 4(p + 8) \times 5(p - 4) } $ $a = \dfrac{ (p - 4)(p + 6)(p + 8)}{ 20(p + 8)(p - 4)} $ Notice that $(p + 8)$ and $(p - 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(p - 4)}(p + 6)(p + 8)}{ 20(p + 8)\cancel{(p - 4)}} $ We are dividing by $p - 4$ , so $p - 4 \neq 0$ Therefore, $p \neq 4$ $a = \dfrac{ \cancel{(p - 4)}(p + 6)\cancel{(p + 8)}}{ 20\cancel{(p + 8)}\cancel{(p - 4)}} $ We are dividing by $p + 8$ , so $p + 8 \neq 0$ Therefore, $p \neq -8$ $a = \dfrac{p + 6}{20} ; \space p \neq 4 ; \space p \neq -8 $